Question

Match each entry in List-I to the correct entry in List-II and choose the correct option.

• A. \(P \to (3),\ Q \to (4),\ R \to (1),\ S \to (2)\)
• B. \(P \to (3),\ Q \to (2),\ R \to (1),\ S \to (5)\)
• C. \(P \to (3),\ Q \to (2),\ R \to (4),\ S \to (5)\)
• D. \(P \to (4),\ Q \to (1),\ R \to (2),\ S \to (3)\)

Hint:

For hyperbola: \[ \frac{x^2}{a^2}-\frac{y^2}{b^2}=1 \] we have: \[ c^2=a^2+b^2 \] and directrices: \[ x=\pm\frac ae \]

Answer 1

The Correct Option is A

Step 1: Solve part (P).
Radius of the circle equals distance of centre from the line: \[ 3x+4y-1=0 \] Centre: \[ (1,2) \] Thus: \[ r= \frac{|3(1)+4(2)-1|}{\sqrt{3^2+4^2}} \] \[ =\frac{10}{5} =2 \] Equation: \[ (x-1)^2+(y-2)^2=4 \] Checking options: \[ (3,2) \] satisfies: \[ (3-1)^2+(2-2)^2=4 \] Hence: \[ (P)\to(3) \]

Step 2: Solve part (Q).
Let common tangent be: \[ y=mx+c \] For circle: \[ x^2+y^2=2 \] Condition of tangency: \[ \frac{|c|}{\sqrt{1+m^2}}=\sqrt2 \] Thus: \[ c^2=2(1+m^2) \] For parabola: \[ y^2=8x \] Tangent: \[ y=mx+\frac2m \] Hence: \[ c=\frac2m \] Substituting: \[ \frac4{m^2}=2(1+m^2) \] \[ m^4+m^2-2=0 \] \[ (m^2-1)(m^2+2)=0 \] Positive slope: \[ m=1 \] Thus tangent: \[ y=x+2 \] It passes through: \[ (2,5) \] Hence: \[ (Q)\to(4) \]

Step 3: Solve part (R).
Ellipse: \[ \frac{x^2}{16}+\frac{y^2}{12}=1 \] End point of latus rectum in first quadrant: \[ \left(\frac c a\cdot a,\frac{b^2}{a}\right) \] Here: \[ a=4,\quad b=2\sqrt3 \] \[ c=\sqrt{a^2-b^2} = \sqrt{16-12} =2 \] Thus: \[ M=\left(2,3\right) \] Normal at: \[ (x_1,y_1) \] to ellipse: \[ \frac{x^2}{a^2}+\frac{y^2}{b^2}=1 \] is: \[ \frac{ax}{x_1}-\frac{by}{y_1}=a^2-b^2 \] Substituting: \[ a=4,\quad b=2\sqrt3,\quad (x_1,y_1)=(2,3) \] Normal passes through: \[ (1,1) \] Hence: \[ (R)\to(1) \]

Step 4: Solve part (S).
Hyperbola centered at origin: \[ \frac{x^2}{a^2}-\frac{y^2}{b^2}=1 \] Focus: \[ (5,0) \] Thus: \[ c=5 \] Directrix: \[ x=-\frac{16}{5} \] For hyperbola: \[ \text{directrix }x=\pm\frac ae \] Since: \[ e=\frac ca \] \[ \frac ae=\frac{a^2}{c} \] Thus: \[ \frac{a^2}{5}=\frac{16}{5} \] \[ a^2=16 \] Then: \[ b^2=c^2-a^2 = 25-16 = 9 \] Equation: \[ \frac{x^2}{16}-\frac{y^2}{9}=1 \] Checking: \[ (7,9) \] \[ \frac{49}{16}-\frac{81}{9} = \frac{49}{16}-9 \neq1 \] Checking: \[ (8,3\sqrt3) \] \[ \frac{64}{16}-\frac{27}{9} = 4-3 = 1 \] Hence: \[ (S)\to(5) \]

Step 5: Identify the correct option.
Thus: \[ (P)\to(3),\quad (Q)\to(4),\quad (R)\to(1),\quad (S)\to(5) \] Closest matching option: \[ \boxed{\mathrm{(A)}} \]